# Locally flat coordinate on Poincaré half plane

I have a question from " Einstein gravity in a nutshell by A. Zee", chapter I.6 question 4:

" Find the locally flat coordinate on the Poincaré half plane."

Also, there is a hint and solution for this question in the back of the book that I copy it:

"With no loss of generality, we can pick the point $$P$$ to be $$(x, y) = (0, y_{\ast})$$. A particular set of locally flat coordinates $$(u, v)$$is given by $$x = y_{\ast}(u + uv + . . .),\,\, y = y_{\ast}(v + \frac{(v^2 − u^2)}{2} + . . .)$$, where the dots represent terms cubic and higher in u, v."

To use the best of your help I am going to show my way of looking at this question.
The metric of the Poincaré half plane is:
$$ds^2= \frac{(dx^2+dy^2)}{y^2}$$ For the flat space the metric: $$ds^2= du^2+dv^2$$ and the metric transformation law is : $$g^{\prime}_{\lambda \sigma}(x^{\prime})=g_{\mu\nu}(x)\frac{\partial{x^{\mu}}}{\partial{{x^{\prime}}^{\lambda}}}\frac{\partial{x^{\nu}}}{\partial{{x^{\prime}}^{\sigma}}}$$ so I have:
$$(\frac{\partial{x}}{\partial{u}})^2+(\frac{\partial{y}}{\partial{u}})^2=y^2 \qquad \qquad (\frac{\partial{y}}{\partial{v}})^2+(\frac{\partial{x}}{\partial{v}})^2=y^2$$ $$\frac{\partial{x}}{\partial{u}}\frac{\partial{x}}{\partial{v}}=-\frac{\partial{y}}{\partial{u}}\frac{\partial{y}}{\partial{v}}$$

Now, if I consider A. Zee's locally flat coordinate at $$(x,y)=(0,y_{\ast})$$ ($$(u,v)=(0,0)$$ in the new coordinate system) with a change (adding $$1$$ in parentheses) :
$$x = y_{\ast}(u + uv),$$
$$y = y_{\ast}(1+v + \frac{(v^2 − u^2)}{2})$$
I will have:
$$\begin{eqnarray} \frac{\partial{x}}{\partial{u}}&=&y_{\ast}(1+v)\qquad \frac{\partial{x}}{\partial{v}}=y_{\ast}u \\ \frac{\partial{y}}{\partial{u}}&=&-y_{\ast}u \qquad \qquad \frac{\partial{y}}{\partial{v}}=y_{\ast}(1+v) \end{eqnarray}$$
From transformation law I have:
$$y^2= y^2_{\ast}(1+v^2+2v+u^2)$$ and $$y^2_{\ast}(1+v)u = y^2_{\ast}(1+v)u$$ which leads to: $$ds^2= du^2+dv^2$$

It seems that if I change $$v \to v+a$$ and the same for u, the transformation law remains unchanged, so the new coordinate system is not unique and the A. Zee's locally flat coordinate is correct.
New update :
I can not solve this problem generally. I would like to derive some higher-order terms. I took $$x= f(u,v)$$ and $$y=g(u,v)$$ and expanded them in u and v using transformation laws, but I could not solve the equations manually.

• what about non-diagonal element of the metric transformation law (i.e. $\sigma\neq\lambda$)? I don't think your solution is compatible with it.
– Kosm
May 11 at 21:13
• @Kosm Your comment was more than useful for me. May I know if you see any mistake in my way?
– Ali
May 12 at 18:25
• I should add that I have changed my solution with the help of the comment that I have received.
– Ali
May 12 at 18:44
• I think the way to find the solution without knowing its form is to expand the equations in $u,v$ up to cubic terms.
– Kosm
May 12 at 19:00
• @Kosm, I could not solve it generally although tried. It is not easy work.
– Ali
May 13 at 19:39